The binding energy of deuteron is `2.2` MeV and that of `._(2)^(4)He` is `28` MeV. If two deuterons are fused to form one `._(2)^(4)He`, th `n` the energy released is

Options

A. `25.8` MeV
B. `23.6` MeV
C. `19.2` MeV
D. `30.2` MeV

Answer

To solve the problem of finding the energy released when two deuterons fuse to form one helium-4 nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Binding Energies**:
- The binding energy of deuteron (\( ^2_1H \)) is given as \( 2.2 \) MeV.
- The binding energy of helium-4 (\( ^4_2He \)) is given as \( 28 \) MeV.

2. **Calculate Total Binding Energy of Two Deuterons**:
- Since we have two deuterons, the total binding energy for the two deuterons is:
\[ \text{Total Binding Energy of 2 Deuterons} = 2 \times 2.2 \text{ MeV} = 4.4 \text{ MeV} \]

3. **Calculate the Energy Released During Fusion**:
- The energy released during the fusion process can be calculated by finding the difference between the binding energy of the helium nucleus and the total binding energy of the two deuterons:
\[ \text{Energy Released} = \text{Binding Energy of Helium} - \text{Total Binding Energy of 2 Deuterons} \]
\[ \text{Energy Released} = 28 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV} \]

4. **Conclusion**:
- Therefore, the energy released when two deuterons fuse to form one helium-4 nucleus is \( 23.6 \) MeV.

### Final Answer:
The energy released is \( 23.6 \) MeV.

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